答案:abc≤1/27

7.已知a>0,b>0,c>0,且a,b,c不全相等,求证:bc/a+ac/b+ab/c>a+b+c.

分析利用综合法证明,注意条件a,b,c不全相等的使用.

证明∵a>0,b>0,c>0,

　　∴bc/a+ac/b≥2√(bc/a "·" ac/b)=2c,

　　bc/a+ab/c≥2√(bc/a "·" ab/c)=2b,

　　ac/b+ab/c≥2√(ac/b "·" ab/c)=2a.

　　又∵a,b,c不全相等,

　　∴上面三式"="号不能同时成立,

　　∴三式相加后两边除以2,得bc/a+ac/b+ab/c>a+b+c.

8.已知an=4n-2n,Tn=2^n/(a_1+a_2+"..." +a_n ),求证:T1+T2+T3+...+Tn<3/2.

证明∵a1+a2+...+an=41+42+43+...+4n-(21+22+...+2n)=(4"(" 1"-" 4^n ")" )/(1"-" 4)-(2"(" 1"-" 2^n ")" )/(1"-" 2)=4/3(4n-1)+2(1-2n),

　　∴Tn=2^n/(4/3 "(" 4^n "-" 1")" +2"(" 1"-" 2^n ")" )=2^n/(4^(n+1)/3 "-" 4/3+2"-" 2^(n+1) )=2^n/(4^(n+1)/3+2/3 "-" 2^(n+1) )=(3"·" 2^n)/(4^(n+1) "-" 3"·" 2^(n+1)+2)=3/2·2^n/(2"·(" 2^n ")" ^2 "-" 3"·" 2^n+1)=3/2·2^n/("(" 2"·" 2^n "-" 1")(" 2^n "-" 1")" )=3/2 (1/(2^n "-" 1) "-" 1/(2^(n+1) "-" 1)).

　　∴T1+T2+T3+...+Tn

　　=3/2 (1"-" 1/3+1/3 "-" 1/7+"..." +1/(2^n "-" 1) "-" 1/(2^(n+1) "-" 1))<3/2.

9.已知x,y,z不全为零,求证:√(x^2+xy+y^2 )+√(y^2+yz+z^2 )+√(z^2+zx+x^2 )>3/2(x+y+z).

证明∵√(x^2+xy+y^2 )=√((x+y/2)^2+3/4 y^2 )≥√((x+y/2)^2 )=|x+y/2|≥x+y/2,

　　同理可得√(y^2+yz+z^2 )≥y+z/2,

　　√(z^2+zx+x^2 )≥z+x/2.

　　又x,y,z不全为零,

　　∴上述三个不等式不能同时取"="号,

∴√(x^2+xy+y^2 )+√(y^2+yz+z^2 )+√(z^2+zx+x^2 )>x+y+z+x/2+y/2+z/2=3/2(x+y+z).